Thursday April 11 Solutions Parametrizations and Integrals 1 Consider the ellipsoid with implicit equation x2 y 2 z2 1 a2 b2 c 2 a Parametrize this ellipsoid Solution One could use the parametrization x a sin cos y b sin sin z c cos 0 0 2 b Set up but do not evaluate a double integral that computes its surface area Solution Since r a sin cos b sin sin c cos one has r a cos cos b cos sin c sin r a sin sin b sin cos 0 so r r bc sin2 cos ac sin2 sin ab sin cos Therefore q r r b 2 c 2 sin4 cos2 a 2 c 2 sin4 sin2 a 2 b 2 sin2 cos2 and the surface area is computed by Z Area 0 Z 0 2 Z 0 r r d d 2 Z q 0 b 2 c 2 sin4 cos2 a 2 c 2 sin4 sin2 a 2 b 2 sin2 cos2 d d 2 Let r u v 2 cos u cos v 2 cos u sin v sin u where 0 u 2 and 0 v 2 a Sketch the surface parametrized by this function Solution The sketch of the surface is as follows b Compute its surface area Solution By the parametrization one has ru sin u cos v sin u sin v cos u rv 2 cos u sin v 2 cos u cos v 0 and so ru rv 2 cos u cos u cos v 2 cos u cos u sin v 2 cos u sin u Therefore ru rv 2 cos u and the surface area is computed by Z Area 2 Z 2 0 0 Z ru rv d ud v 2 Z 2 0 0 2 cos u d ud v 8 2 3 Consider the surface integral z dS where is the surface with sides S 1 given by the cylinder x 2 y 2 1 S 2 given by the unit disk in the x y plane and S 3 given by the plane z x 1 Evaluate this integral as follows a Parametrize S 1 using z coordinates Solution One can parametrize S 1 by x cos y sin z z 0 2 0 z cos 1 b Evaluate the integral over the surface S 2 without parametrizing Solution Since z 0 on S 2 we know S2 z d S 0 c Parametrize S 3 in Cartesian coordinates and evaluate the resulting integral using polar coordinates Solution One can parametrize S 3 in Cartesian coordinates p p x x y y z x 1 1 x 1 1 x 2 y 1 x 2 Now we move to evaluate the integral z d S Obviously z dS z dS z dS z d S I 1 I 2 I 3 S1 S2 S3 To estimate I 1 using the parametrization in a one has r z cos sin z Then r sin cos 0 rz 0 0 1 and r rz cos sin 0 So r rz 1 and 2 Z cos 1 cos 1 2 d 2 0 0 0 Z 2 3 cos2 2 cos 1 d 2 2 0 Z I1 z d zd Z 2 In b we know I 2 0 To evaluate I 3 by the parametrization in c one has p p r x y x y x 1 1 x 1 1 x 2 y 1 x 2 and so rx 1 0 1 r y 0 1 0 p Thus rx r y 2 and the surface integral is Z I3 1 Z p 1 x 2 p 1 1 x 2 rx r y 1 0 1 p x 1 2 d yd x x 2 y 2 1 p x 1 2 d yd x To evaluate this integral one can use the polar coordinates x r cos y r sin 0 r 1 0 2 Therefore 2 Z 1 Z I3 0 0 p p r cos 1 2 r d r d 2 Adding up all three integrals one gets 3 p z d S I1 I2 I3 2 2 4 Let C be the circle in the plane with equation x 2 y 2 2x 0 a Parametrize C as follows For each choice of a slope t consider the line L t whose equation is y t x Then the intersection L t C of L t and C contains two points one of which is 0 0 Find the other point of intersection and call its x and y coordinates x t and y t Compute a formula for r t x t y t Check your answer with your TA Solution Bring y t x into x 2 y 2 2x 0 then one has x 2 t 2 x 2 2x 0 and it is 2 2t 2 2t easy to get x 1 t 2 and then y 1 t 2 Thus r t 1 t 2 1 t 2 p is a rational number Show that x p q and y p q are also rational q numbers Explain how by clearing denominators in x p q 1 and y p q you can find a a triple of integers U V and W for which U 2 V 2 W 2 b Suppose that t Solution Plug t p q into the the parametrization one gets 2q 2 x p q 2 p q2 y p q 2pq p2 q2 and both of them are rational numbers Since x 1 2 y 2 1 and x p q 1 one has 2 2 2pq 2 q p2 2 1 p2 q2 p q2 By setting U q 2 p 2 V 2pq W p 2 q 2 one has U 2 V 2 W 2 Z c Compute C 1 y x d r using your parametrization above 2 2 2 2t 4t 2 2t 0 Solution Since r 1 t 2 1 t 2 one has r 1 t 2 2 1 t 2 2 Then Z C 1 2t 2 4t 2 2t 2 dt 2 1 t2 1 t2 1 t 2 2 1 t 2 2 Z 2 d t 2 2 1 t 1 y x d r 2 Z q 2 p 2 then p 2 q 2
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